Current density, drift velocity and electric field

[Mauvai]

Homework Statement

A) A current I flows along a straight wire of cross-sectional area A. If the unit vector I denotes the direction of current flow, write down the expression for the current density J (assuming it to be uniform throughout the wire ) and give its SI units.

B) (This isn't really relevant) The expression for J obtained in part (a) is in terms of the macroscopic quantities I and A. Now derive the corresponding expression in terms of microscopic quantities, namely J = n q vd, where n is the number of charge carriers per unit volume, q is the charge carried by each charge carrier, and vd is the average drift velocity of the charge carriers.
Verify that the SI units of n q v_d agree with the units for ~J in part (a)

C) A current of 0.5 A flows in a copper wire of cross-sectional area 0.3 mm2.
(i) Calculate the current density in the wire.
(ii) Calculate the average drift speed of the charge carriers.
(iii) Calculate the electric eld strength in the wire using the
formula E =  ρJ where ρ is the resistivity.

Homework Equations

J = I/A
J = nqv_d
E = ρJ

The Attempt at a Solution

First few bits are easy - current density is I/A, units A/m²

Proof is standard: I/A = nqV/A where V is volume, n is charges per volume.
V/A = Δx, and Δx/t is v_d ⇔ J = nqv_d

c(i) easy, fill in formula - j = 530500 Am^-2
c(ii) this is where the problems start - again its easy, fill in the formula, but i have not been given the value of n (q is standard, 1.602x10^-19)
So my question is, is it possible to find v_d without n? is there an alternate way of working it out?
c(iii) same as c(ii) but with ρ - that wasn't provided either.
Thanks!

 

[TSny]

Proof is standard: I/A = nqV/A where V is volume, n is charges per volume.
V/A = Δx, and Δx/t is v_d ⇔ J = nqv_d

The equation I/A = nqV/A is not correct. There needs to be a time interval t in the denominator on the right-hand side.

c(i) easy, fill in formula - j = 530500 Am^-2

I don't get this result. I get j = I/A = 0.5 A/( 0.3 mm²) = 0.5 A/(0.3 x 10^-6 m²) = 1.7 x 10⁶ A/m²

c(ii) this is where the problems start - again its easy, fill in the formula, but i have not been given the value of n (q is standard, 1.602x10^-19)
So my question is, is it possible to find v_d without n? is there an alternate way of working it out?
c(iii) same as c(ii) but with ρ - that wasn't provided either.
Thanks!

You are probably expected to look up n and ρ for copper. Many textbooks have tables that list these for different materials.